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Three-Phase • 400V • Copper Cable

Calculate Wire Cross-Section for Three-Phase 400V

Calculate the required wire cross-section for a copper cable based on power, cable length, cos φ and permissible voltage drop.

Calculation Values

Note: This calculation is a technically sound simplification typical of online calculators. For practical sizing, standards, installation method, temperature, grouping, protective devices, cable type and manufacturer specifications must also be checked.

Formula Used

For three-phase systems:

I = P / (√3 × U × cos φ)

A = (√3 × L × I × cos φ) / (κ × ΔU)

Here, A is the conductor cross-section in mm², L the one-way cable length in meters, I the current in amps, κ the conductivity of copper, defaulting to 58, and ΔU the permissible voltage drop in volts. The power factor cos φ is properly accounted for in the current and voltage drop calculation.

FAQ

Frequently Asked Questions about Calculate Wire Cross-Section for Three-Phase Copper Cable

Briefly explained: formula, use case and key limits of the calculation.

Which formula does the three-phase copper calculator use? +

The calculator uses I = P / (√3 × U × cos φ) to determine the current and A = (√3 × L × I × cos φ) / (κ × ΔU) for the conductor cross-section. For copper, κ defaults to 58.

Why does the power factor cos φ matter? +

The power factor accounts for the fact that, with inductive or capacitive loads, not all apparent power is converted into real power. For purely resistive loads, cos φ can be set to 1.

Is the calculated cross-section sufficient for an installation? +

No. The calculated value is a guideline based on voltage drop. Current-carrying capacity, installation method, grouping, ambient temperature, protective devices and applicable standards must also be checked.

Why is a standard cross-section recommended? +

The calculated value rarely matches a commercially available conductor cross-section. It is therefore rounded up to the next common standard size.

Practical guidance

Calculate a 400 V copper cable from power and voltage drop

The page derives balanced current and then the minimum cross-section that meets the selected voltage-drop target.

Input values explained

Power in kilowatts
Enter the electrical active power in kW. One kW equals 1,000 W; motor output power and electrical input power are not necessarily the same.
Cable length
Enter the one-way distance from supply point to load. The required return path is already represented by the selected formula.
Voltage
Enter the line-to-line voltage between two phases. This is commonly 400 V in low-voltage systems; 230 V is not the correct input here.
Power factor cos φ
cos φ is the ratio of active to apparent power. It is close to 1 for resistive heating loads; use the data-sheet value for motors and transformers.
Permitted voltage drop
This percentage limits the calculated voltage loss. It is a design assumption and must suit the circuit and applicable requirements.
Conductivity κ
κ is the calculation value for the conductor material. Common approximations are 58 for copper and 37 for aluminium; heating increases actual resistance.

Example: 11 kW over 25 m

At 400 V, 11 kW, cos φ 0.90 and 3%, 12 V of loss is allowed.

I ≈ 17.64 A; A = (√3 × 25 × 17.64 × 0.90) ÷ (58 × 12) ≈ 0.99 mm²

Voltage drop indicates 1.5 mm². Thermal ampacity will often require a larger size for a real 11 kW circuit.

State the result as a voltage-drop minimum

Only comparison with ampacity and protective requirements can produce a defensible final cable size.

Calculation limits

  • Motor starting, harmonics and neutral loading are excluded.
  • Cable type, temperature, installation method and protection are unknown.